I know exactly what your teacher is trying to describe. And your teacher's explanation is rubbish. He is using the marching soldiers or the two wheels hitting the sand analogy you see here.
It doesn't seem like you included the coupling. But, even if you did, I know you could show similar looking equations. These equations would probably not explicitly show that there is a dispersion relation.
I think we diverge on what an uncoupled HO is defined to be. For me, it is something...
You might want to search for "sharp principal diffuse fundamental series equations" or "alkali spectra".
Here, here and https://www.tcd.ie/Physics/people/Peter.Gallagher/lectures/js_atomic/JS_atomic_lecture8_9.pdf might be useful links.
I believe earlier it was implied that there is consensus on this use of the term independent harmonic oscillators in second quantization. Going through many online course notes, I find that this isn't the case. I find that there is a spectrum of opinions that run from i) yes, it is a harmonic...
1) You could try calculating the wavenumbers for both and comparing them to experiment.
2) Eq. 2.15 seems to show you that Eq. 2.12 is a typo.
3) Each of those equations has a ##\beta## with a different subscript, so they are going to give different results.
I think we have more points of agreement than disagreement. I'm not denying the formalism of second quantization. The two systems of coupled and independent oscillators are similar. The Hamiltonians that are typically shown in second quantization sure seem to be similar.
I'm trying to say that...
It would really help if you wrote out here the equations of interest. Having a brief look at your code, I believe you are trying to calculate the eccentricity. My guess is that you are using the equation that only works with ##0<e<1##, which is for an ellipse.
I was confused by his first sentence. It's not correct.
Acoustic phonons are waves that can transport heat. Uncoupled oscillators can't do that. They are different physical systems.
Well now I seem to be confused by your confusion of my confusion.:wink:
The uncoupled Hamiltonian will give the rather unphysical Einstein mode, which is like I said a poor man's model of optical phonons.
The coupled Hamiltonian I've given is for a monatomic chain. It will give only acoustic...
We use the term quasiparticle because one might intuitively think it is impossible to resolve the behaviour of the individual particles separate from all the things they are correlated with. Surprisingly, we can still model solids as if they had things that behave similar to particles that can...
I am very confused by why you are saying this.
Coupled HO's will give you phonon dispersion relations which have a dependence on crystal momentum ##k##. Whenever, I've seen a phonon spectrum modeled or measured there is a cap to the maximum energy.
An array of uncoupled HO's are what we would...
If I move my phone near the back of the TV I get about 150 microTesla.
I'll tell you that I've been regularly bathed in the stray field of a 9 Tesla magnet with no ill effects. This field is strong enough to pick up nearby metal objects or to force my phone to shut down when I left it 2.5 m...
No. You've been doing it right. Why are you trying to change things? Your black line to fits well with the coloured lines. You are done. There is nothing more to discuss here.
You want to go further into the region where the data curves and was caused by experimental error? The definition of Young's modulus requires a linear regime.
Have a closer look at the final image you shared. All the slopes are extremely similar. Don't you think that your Young modulus numbers should be very close? One of your values is far off because the software is giving you an erroneous number.
Gaseous atoms, ions, and molecules all have discrete emission spectra. The excitation doesn't need to be from photons; it can also be from bombardment with electrons (in fluorescent tubes electron bombard mercury). There are also often multiple decay pathways so you get out a multiple...
What do you mean by why the sun stays on course?
Two more helpful things to look at are the tropical year and the apparent solar time (regarding a sundial).
I've never done the derivation myself. My suggestion is to first think of what will happen with one reflected wave. Draw it out and then write down an equation. Then do it for the second reflection. I think eventually you will see the pattern.
You might also pulling out terms from your...
The image you have is of a thing called a Fabry-Perot etalon.
In reality, to calculate the phase change you need to do a Kramers-Kronig integral of reflectivity from 0 to infinite frequency. Obviously, they didn't give you that information and the math is for experts. We usually just say that...
From my reading on antiquity and their cosmological views, I learned that in approximately 70 years the zodiacs shift about 1 degree.
Well I just looked it up. It's actually 71.39 years for one degree of precession.
I've checked out the software. It is definitely buggy. If you move the cursor (black crosshairs) horizontally, you can see the strain jump to numbers it shouldn't. If you move the cursor vertically, you can see it jump to numbers you shouldn't.
Luckily, in the region that you need to measure...
To expand on @Ibix 's good answer: your brain can be stimulated to perceive green by having photons of around 440 nm be absorbed by your cones, or have only two wavelengths corresponding to red and blue photons be absorbed in the right ratio on the same area of your retina.
Am I correct in reading that in this merged thread that post #1 is about a spinning gyroscope and post #2, while mentioning your want to understand gyroscopes, is actually about a rod pivoting downward?
I read it differently the first time I saw your post before today because of your mention of...
Why do you think the normal force at the pivot is equal to the weight? You have a contradiction here, don't you? It looks like you've started with Newton's second law:
$$\Sigma F_y = ma_y,$$
and jumped right to setting the acceleration to zero.
From walking my dog at night, I learned that I can tell from a distance whether the colour from xmas lights are made from two different LEDs or not, or whether a car's brake lights are incandescent or an LED. I think it has to do with persistence of vision and the pulse rate of LEDs.
I just...
I like to show my students a visible spectrum and ask them to point out where the color white is. Then I show them that their minds are tricking themselves(?) by using a handheld spectroscope and looking at the emission lines from white fluorescent lights.
It really seems to be that the numbers they give you are wrong. This is not unheard of with these types of programs. Can you share an image of a different graph? Maybe, another one will have correct values.
If you fix the factor of 10 that @haruspex mentioned then you are doing the calculation correctly. However, I am unsure if this software is giving you the correct values. You should be getting an answer close to 200 GPa for steels. Your black crosshair doesn't match with the stress axis, but...
Have you seen the demonstration of precession of a bicycle wheel from a string? I think this is the easiest way to understand:
The wheel is spinning and hung on the string. Let's just consider two infinitesimal points on the rim of the wheel at the top and the bottom at some instant. I guess...
There are many detectors around the world trying to detect dark matter particles directly via some other assumed non-gravitational interaction. So far the evidence points to dark matter particles interacting only via gravity.
On the other hand, there are cosmological simulations that have...
I believe that Neil deGrasse Tyson's hidden premise in that statement would be something like: the observations that we label as being due to dark matter are due to matter particles. Thus the models he considers exclude things like modified Newtonian dynamics (MOND).
The term dark matter is...
Here's the first two hints:
If a particle moves directly at another particle, mathematically how is this represented?
You'll also need to write out an equation for the motion of particle 1, which will be used later.
Just a point of vocabulary: the toe region is the region at the beginning from 0 to approximately 1.5E-5.
To find the slope of the line you need two pairs of points. One pair is often (0, 0) so you may have overlooked this. Look at the image here.
Thinking classically in a quantum mechanical problem is usually not the way to go. That being said: classically, the only way any of us can be unbound from Earth's gravitational well is to be at the escape velocity at Earth's surface. We need the kinetic energy.
Well, you still need to conserve...
This is not my field, but I don't think alpha decay is a general method for nucleons to tunnel out. The isotopes that we encounter on earth, yes, have alpha decay as the dominant way that nucleons tunnel out. However, we see other dominant decay mechanisms in more unstable isotopes.
I think you need to compare apples to apples. Count up the same number of energy levels in each nucleon's well and look at the energy difference. For example, you could use the first level. The proton in the first level is at a higher energy than a neutron at the first level. This shifted energy...
Definitely try to have fun with the math. Sometimes ditch the textbook for a little while and play with math you've already learned. Try to make up a problem for you to solve. See what is and isn't possible.
Math is also more than just symbols. In a spreadsheet, you can play around with...
I still can't picture the motion. An object can travel in a circle with constant speed, and if you take one component of its motion then you will have a one-dimensional oscillator. However, if you have a one-dimensional oscillator it must turn around 180 degrees at its maximum displacement. To...